Velocity is the time derivative of position, and by that token position can be found by integrating a known velocity function with respect to time:

\(\displaystyle p(t)=\int v(t)dt\)

Now, if this integral were to be taken over an interval of time \(\displaystyle t=[a,b]\), this will give a finite value, a change in position, i.e. a distance travelled:

\(\displaystyle d=\int_a^b v(t)dt\)

For the velocity function

\(\displaystyle v(t)=8sin(t)cos(t)\)

The distance travelled can be found via knowledge of the following derivative properties:

Trigonometric derivative: 

\(\displaystyle d[sin(u)]=cos(u)du\)

\(\displaystyle w=sin(t);dw=cos(t)\)

\(\displaystyle \int8wdw=4w^2\)

The distance travelled over \(\displaystyle t=[0,\frac{\pi}{2}]\) is:

 \(\displaystyle d=4sin^2(t)|_0^{\frac{\pi}{2}}\)

\(\displaystyle d=(4sin^2(\frac{\pi}{2}))-4sin^2(0)=4\)

This is simply the application of the distance formula:

The distance \(\displaystyle d\) is going to be equal to:

\(\displaystyle d=\sqrt{(y_1-y_0)^2+(x_1-x_0)^2}=\sqrt{(-4+5)^2+(5+3)^2}=\sqrt{65}\)

We know that displacement is just the magnitude of the vector moving from point \(\displaystyle a\) to point \(\displaystyle b\). Distance however is the sum of the movements. In this question, we can treat the distance as the hypotenuse of a triangle, and the movements in the west and north directions as the 2 legs. Since the person moved the same distance in each direction, which I will call \(\displaystyle a\), we can determine it by doing:

\(\displaystyle 100^2= a^2+a^2\)

\(\displaystyle \sqrt{\frac{100^2}{2}}=50\sqrt{2}=a\)

Since the person walked \(\displaystyle a\) in both directions, the total distance travelled is:

\(\displaystyle 50\sqrt{2}+50\sqrt{2}=100\sqrt{2}=a\)

The displacement in orthogonal directions(North and West) can be determined by using the pythagorean theorem:

\(\displaystyle d=\sqrt{30^2+70^2}=\sqrt{5800}=10\sqrt{58}\)

In general, if we have a vector (a,b), a perpendicular vector is (b,-a).

So here, the perpendicular vector is (3,-4).

To express the vector in terms of i, j, and k, we need to combine like terms and distribute.

3a + 2b

= 3(i + 2j - 3k) + 2(4i + 7k)

= 3i + 6j - 9k + 8i + 14k

= 11i +6j + 5k

To find the position from velocity, the function must be integrated.  This gives \(\displaystyle s(t)=2t^{3/2}+t^2+C\).  substituting 4 for \(\displaystyle t\) and using the given initial condition gives the answer

The position function is the intergral of the velocity function.  So here, position is given by \(\displaystyle s = 3t^2+C\) where \(\displaystyle C\) is the constant of integration.  Because only a difference in position is asked, and not an absolute position, the constant of integration cancels out.  

To find the position from the acceleration function, integrate the acceleration function twice.

\(\displaystyle v(t)= \int a(t)dt = t\)

\(\displaystyle s(t) = \int v(t) dt = \frac{t^2}{2}\)

Substitute \(\displaystyle t=2\) to find the postion.

\(\displaystyle s(t=2)= \frac{2^2}{2} = 2\)

To find the position function, integrate the acceleration function twice.

\(\displaystyle v(t)= \int a(t) dt= \int (t^2+2t+1 )dt = \frac{1}{3}t^3 + t^2 +t\)

\(\displaystyle s(t)=\int v(t)dt=\int (\frac{1}{3}t^3+t^2+t) \:dt=\frac{1}{12}t^4+\frac{1}{3}t^3+\frac{1}{2}t^2\)

Evalute the position at \(\displaystyle t=2\).

\(\displaystyle s(t=2)=\frac{1}{12}(2)^4+\frac{1}{3}(2)^3+\frac{1}{2}(2)^2=\frac{4}{3}+\frac{8}{3}+2=6\)